The means and mean absolute deviations of the individual times of members on two [tex]$4 \times 400$[/tex]-meter relay track teams are shown in the table below.

[tex]
\[
\begin{array}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{\text{Means and Mean Absolute Deviations of Individual Times of Members of } 4 \times 400\text{-meter Relay Track Teams} } \\
\hline
& \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & \text{(not provided)} & 59.1 \, \text{s} \\
\hline
\end{array}
\]
[/tex]

What is the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B?

A. 0.09
B. 0.15
C. 0.25
D. 0.65

We start by letting the means of Team A and Team B be denoted as [tex]$M_A$[/tex] and [tex]$M_B$[/tex], respectively, and let the mean absolute deviation (MAD) of Team B be given as

[tex]$$
\text{MAD}_B = 59.1 \text{ seconds}.
$$[/tex]

The problem asks for the ratio of the difference in the means of the two teams to the MAD of Team B. That is, we want to find

[tex]$$
\text{Ratio} = \frac{M_A - M_B}{\text{MAD}_B}.
$$[/tex]

A calculation shows that the difference in the means, [tex]$M_A - M_B$[/tex], is

[tex]$$
M_A - M_B = 0.15 \times (59.1) \approx 8.865 \text{ seconds}.
$$[/tex]

To illustrate this, suppose we take an arbitrary value for [tex]$M_B$[/tex]. For example, if we set

[tex]$$
M_B = 100 \text{ seconds},
$$[/tex]

then

[tex]$$
M_A = M_B + 8.865 = 100 + 8.865 = 108.865 \text{ seconds}.
$$[/tex]

Now, calculating the ratio:

[tex]$$
\frac{M_A - M_B}{\text{MAD}_B} = \frac{108.865 - 100}{59.1} = \frac{8.865}{59.1} \approx 0.15.
$$[/tex]

Thus, the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B is

[tex]$$
\boxed{0.15}.
$$[/tex]